數學功課.要有步驟ar...help ME..急!!

🎓 學業台小學及中學教育

Man Kwan

2007-11-07 4:13 am

要有步驟ar..

1.(6a+5)+(2a-4)
=
=
=
=

2.(y^2+3)+(2y^2-7)
=
=
=
=

3.(2d-3d^2)-(d^2+3d)
=
=
=
=
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回答 (2)

白奇偉

2007-11-07 4:26 am

✔ 最佳答案

1.(6a+5)+(2a-4)
=6a+2a+5-4
=8a+1

2.(y²+3)+(2y²-7)
=y²+2y²+3-7
=3y²-4

3.(2d-3d²)-(d²+3d)
=2d-3d²-d²-3d
=2d-3d-3d²-d²
=-d-4d²

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用戶73615

2007-11-07 4:43 am

1.(6a+5)+(2a-4)
=6a+5+2a-4
=8a+1

2..(y^2+3)+(2y^2-7)
=y^2+3+2y^2-7
=3y^2-4

3.(2d-3d^2)-(d^2+3d)
=.2d-3d^2-d^2-3d
=-3d^2-d^2-3d+2d
=-4d^2-d

參考: me

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